Approximate $\sqrt{(1.02)^3+(1.97)^3}$ using differential

Question

So $$\sqrt{(1.02)^3+(1.97)^3}=\sqrt{(1+0.02)^3+(2-0.03)^3}$$

So the differential will be $$\sqrt{1^3+2^3}+\frac{3(1)^2}{2\sqrt{1^3+2^3}}\Delta x+\frac{3(2)^2}{2\sqrt{1^3+2^3}}\Delta y=3+\frac{1}{2}\Delta x+2\Delta y$$

is $\Delta x=0.02 \text{ and } \Delta y=-0.03$ or $\Delta x=(0.02)^3 \text{ and } \Delta y=(-0.03)^3$?

Answer 1

For $$\sqrt{(1.02)^3+(1.97)^3}=\sqrt{(1+0.02)^3+(2-0.03)^3}$$

The correct choice is $$\Delta x=0.02 \text{ and } \Delta y=-0.03$$

Note that you have changed your $x$ from $1$ to $1.02$, which means your change in $x$ is $ \Delta x=0.02$

Similarly for change in $y$ from $2$ to $1.97$ we have $\Delta y=-0.03$

Answer 2

Note that $$\sqrt{1.02^3+1.97^3}=\sqrt{(x+\Delta x)^3+(y+\Delta y)^3}$$ where $\Delta x=0.02$ and $\Delta y= -0.03$ with $x=1$ and $y=2$.